The movie Moneyball focuses on the “quest for the secret of success in baseball”. It follows a low-budget team, the Oakland Athletics, who believed that underused statistics, such as a player’s ability to get on base, better predict the ability to score runs than typical statistics like home runs, RBIs (runs batted in), and batting average. Obtaining players who excelled in these underused statistics turned out to be much more affordable for the team.
In this lab we’ll be looking at data from all 30 Major League Baseball teams and examining the linear relationship between runs scored in a season and a number of other player statistics. Our aim will be to summarize these relationships both graphically and numerically in order to find which variable, if any, helps us best predict a team’s runs scored in a season.
Let’s load up the data for the 2011 season.
load("more/mlb11.RData")
In addition to runs scored, there are seven traditionally used variables in the data set: at-bats, hits, home runs, batting average, strikeouts, stolen bases, and wins. There are also three newer variables: on-base percentage, slugging percentage, and on-base plus slugging. For the first portion of the analysis we’ll consider the seven traditional variables. At the end of the lab, you’ll work with the newer variables on your own.
runs
and one of the other numerical variables? Plot this relationship using the variable at_bats
as the predictor. Does the relationship look linear? If you knew a team’s at_bats
, would you be comfortable using a linear model to predict the number of runs?I would use a scatterplot.
plot(mlb11$at_bats, mlb11$runs)
The relationship bewteen at bats and runs looks linear. In order to use a linear model to predict the number of runs, I would need to ensure that the relationship between the variables looks linear, the residuals are nearly normal, that there is constant variability, and that the observations are independent.
If the relationship looks linear, we can quantify the strength of the relationship with the correlation coefficient.
cor(mlb11$runs, mlb11$at_bats)
## [1] 0.610627
Think back to the way that we described the distribution of a single variable. Recall that we discussed characteristics such as center, spread, and shape. It’s also useful to be able to describe the relationship of two numerical variables, such as runs
and at_bats
above.
The relationship between at bats and runs is positive and linear. There is one point that is far from the rest of the data that corresponds to a large number of times at bat. Because it is far from the rest of the data, it will have a large affect on the linear regression line. There are 3 points that represent players with a large number of runs. The amount of linearity is not very strong.
Just as we used the mean and standard deviation to summarize a single variable, we can summarize the relationship between these two variables by finding the line that best follows their association. Use the following interactive function to select the line that you think does the best job of going through the cloud of points.
plot_ss(x = mlb11$at_bats, y = mlb11$runs)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## -2789.2429 0.6305
##
## Sum of Squares: 123721.9
After running this command, you’ll be prompted to click two points on the plot to define a line. Once you’ve done that, the line you specified will be shown in black and the residuals in blue. Note that there are 30 residuals, one for each of the 30 observations. Recall that the residuals are the difference between the observed values and the values predicted by the line:
\[ e_i = y_i - \hat{y}_i \]
The most common way to do linear regression is to select the line that minimizes the sum of squared residuals. To visualize the squared residuals, you can rerun the plot command and add the argument showSquares = TRUE
.
plot_ss(x = mlb11$at_bats, y = mlb11$runs, showSquares = TRUE)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## -2789.2429 0.6305
##
## Sum of Squares: 123721.9
Note that the output from the plot_ss
function provides you with the slope and intercept of your line as well as the sum of squares.
plot_ss
, choose a line that does a good job of minimizing the sum of squares. Run the function several times. What was the smallest sum of squares that you got? How does it compare to your neighbors?The smallest sum of squares value I got when testing it myself is 123773.4
It is rather cumbersome to try to get the correct least squares line, i.e. the line that minimizes the sum of squared residuals, through trial and error. Instead we can use the lm
function in R to fit the linear model (a.k.a. regression line).
m1 <- lm(runs ~ at_bats, data = mlb11)
The first argument in the function lm
is a formula that takes the form y ~ x
. Here it can be read that we want to make a linear model of runs
as a function of at_bats
. The second argument specifies that R should look in the mlb11
data frame to find the runs
and at_bats
variables.
The output of lm
is an object that contains all of the information we need about the linear model that was just fit. We can access this information using the summary function.
summary(m1)
##
## Call:
## lm(formula = runs ~ at_bats, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -125.58 -47.05 -16.59 54.40 176.87
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2789.2429 853.6957 -3.267 0.002871 **
## at_bats 0.6305 0.1545 4.080 0.000339 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 66.47 on 28 degrees of freedom
## Multiple R-squared: 0.3729, Adjusted R-squared: 0.3505
## F-statistic: 16.65 on 1 and 28 DF, p-value: 0.0003388
Let’s consider this output piece by piece. First, the formula used to describe the model is shown at the top. After the formula you find the five-number summary of the residuals. The “Coefficients” table shown next is key; its first column displays the linear model’s y-intercept and the coefficient of at_bats
. With this table, we can write down the least squares regression line for the linear model:
\[ \hat{y} = -2789.2429 + 0.6305 * atbats \]
One last piece of information we will discuss from the summary output is the Multiple R-squared, or more simply, \(R^2\). The \(R^2\) value represents the proportion of variability in the response variable that is explained by the explanatory variable. For this model, 37.3% of the variability in runs is explained by at-bats.
homeruns
to predict runs
. Using the estimates from the R output, write the equation of the regression line. What does the slope tell us in the context of the relationship between success of a team and its home runs?plot_ss(x = mlb11$homeruns, y = mlb11$runs, showSquares = TRUE)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## 415.239 1.835
##
## Sum of Squares: 73671.99
hmruns <- lm(runs ~ homeruns, data=mlb11)
summary(hmruns)
##
## Call:
## lm(formula = runs ~ homeruns, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -91.615 -33.410 3.231 24.292 104.631
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 415.2389 41.6779 9.963 1.04e-10 ***
## homeruns 1.8345 0.2677 6.854 1.90e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 51.29 on 28 degrees of freedom
## Multiple R-squared: 0.6266, Adjusted R-squared: 0.6132
## F-statistic: 46.98 on 1 and 28 DF, p-value: 1.9e-07
runs = 415.2389 + 1.8345 * homeruns
The slope tells us that an increase in 1 homerun corresponds to an increase of 1.8345 runs.
Let’s create a scatterplot with the least squares line laid on top.
plot(mlb11$runs ~ mlb11$at_bats)
abline(m1)
The function abline
plots a line based on its slope and intercept. Here, we used a shortcut by providing the model m1
, which contains both parameter estimates. This line can be used to predict \(y\) at any value of \(x\). When predictions are made for values of \(x\) that are beyond the range of the observed data, it is referred to as extrapolation and is not usually recommended. However, predictions made within the range of the data are more reliable. They’re also used to compute the residuals.
library(dplyr)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
library(tidyr)
newdata = data.frame(at_bats=5579)
predict(m1, newdata, interval="predict")
## fit lwr upr
## 1 728.5955 589.0707 868.1204
mlb11 %>%
filter(at_bats==5579) %>%
select(runs, at_bats)
## runs at_bats
## 1 713 5579
Since there was no team with 5,578 at-bats, I am going to look for the residual for 5,579 at-bats. A team with 5,579 at-bats, had 713 runs. The predicted value from the linear regression is 728.6 runs. The residual is 713 - 728.6 = -15.6. The linear regression over-estimated the value.
To assess whether the linear model is reliable, we need to check for (1) linearity, (2) nearly normal residuals, and (3) constant variability.
Linearity: You already checked if the relationship between runs and at-bats is linear using a scatterplot. We should also verify this condition with a plot of the residuals vs. at-bats. Recall that any code following a # is intended to be a comment that helps understand the code but is ignored by R.
plot(m1$residuals ~ mlb11$at_bats)
abline(h = 0, lty = 3) # adds a horizontal dashed line at y = 0
There is no apparent pattern in the residuals plot. This indicates that the relationship between runs and at-bats is linear.
Nearly normal residuals: To check this condition, we can look at a histogram
hist(m1$residuals)
or a normal probability plot of the residuals.
qqnorm(m1$residuals)
qqline(m1$residuals) # adds diagonal line to the normal prob plot
Based on the histogram and the normal probability plot, the nearly normal residuals condition does appear to be met.
Constant variability:
The spacing of the points away from the linear regression line is fairly consistent. The constant variability condition appears to be met.
* * *
mlb11
that you think might be a good predictor of runs
. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?I will test the relationship between the number of stolen bases and runs.
plot(mlb11$stolen_bases, mlb11$runs, xlab = "Stolen bases", ylab = "Runs")
stlnbas <- lm(runs ~ stolen_bases, data=mlb11)
abline(stlnbas)
There does not seem to be a relationship between stolen bases and runs. If there is a relationship between stolen bases and runs, it is a weak one.
runs
and at_bats
? Use the R\(^2\) values from the two model summaries to compare. Does your variable seem to predict runs
better than at_bats
? How can you tell?summary(stlnbas)
##
## Call:
## lm(formula = runs ~ stolen_bases, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -139.94 -62.87 10.01 38.54 182.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 677.3074 58.9751 11.485 4.17e-12 ***
## stolen_bases 0.1491 0.5211 0.286 0.777
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 83.82 on 28 degrees of freedom
## Multiple R-squared: 0.002914, Adjusted R-squared: -0.0327
## F-statistic: 0.08183 on 1 and 28 DF, p-value: 0.7769
The R\(^2\) value is 0.00291. That indicates no relationship between stolen bases and runs. .291% of the variability in runs is due to stolen bases. The R\(^2\) value to predict runs from at-bats is .3739. 37% of the variability in runs is due to at-bats. The stolen bases variable is a terrible predictor for runs.
runs
and each of the other five traditional variables. Which variable best predicts runs
? Support your conclusion using the graphical and numerical methods we’ve discussed (for the sake of conciseness, only include output for the best variable, not all five).plot(mlb11$bat_avg, mlb11$runs, xlab = "Stolen bases", ylab = "Runs")
batavg <- lm(runs ~ bat_avg, data=mlb11)
abline(batavg)
summary(batavg)
##
## Call:
## lm(formula = runs ~ bat_avg, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -94.676 -26.303 -5.496 28.482 131.113
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -642.8 183.1 -3.511 0.00153 **
## bat_avg 5242.2 717.3 7.308 5.88e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 49.23 on 28 degrees of freedom
## Multiple R-squared: 0.6561, Adjusted R-squared: 0.6438
## F-statistic: 53.41 on 1 and 28 DF, p-value: 5.877e-08
Batting average is the best predictor for runs. There is an R\(^2\) value of .6561. That means that 65.61% of the variability in runs is due to batting average.
runs
? Using the limited (or not so limited) information you know about these baseball statistics, does your result make sense?plot(mlb11$new_onbase, mlb11$runs, xlab = "On Base Percentage", ylab = "Runs")
nwbas <- lm(runs ~ new_onbase, data=mlb11)
abline(nwbas)
summary(nwbas)
##
## Call:
## lm(formula = runs ~ new_onbase, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -58.270 -18.335 3.249 19.520 69.002
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1118.4 144.5 -7.741 1.97e-08 ***
## new_onbase 5654.3 450.5 12.552 5.12e-13 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 32.61 on 28 degrees of freedom
## Multiple R-squared: 0.8491, Adjusted R-squared: 0.8437
## F-statistic: 157.6 on 1 and 28 DF, p-value: 5.116e-13
plot(mlb11$new_slug, mlb11$runs, xlab = "Slugging Percentage", ylab = "Runs")
slug <- lm(runs ~ new_slug, data=mlb11)
abline(slug)
summary(slug)
##
## Call:
## lm(formula = runs ~ new_slug, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -45.41 -18.66 -0.91 16.29 52.29
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -375.80 68.71 -5.47 7.70e-06 ***
## new_slug 2681.33 171.83 15.61 2.42e-15 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 26.96 on 28 degrees of freedom
## Multiple R-squared: 0.8969, Adjusted R-squared: 0.8932
## F-statistic: 243.5 on 1 and 28 DF, p-value: 2.42e-15
plot(mlb11$new_obs, mlb11$runs, xlab = "On Base and Slugging Percentage", ylab = "Runs")
basslug <- lm(runs ~ new_obs, data=mlb11)
abline(basslug)
summary(basslug)
##
## Call:
## lm(formula = runs ~ new_obs, data = mlb11)
##
## Residuals:
## Min 1Q Median 3Q Max
## -43.456 -13.690 1.165 13.935 41.156
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -686.61 68.93 -9.962 1.05e-10 ***
## new_obs 1919.36 95.70 20.057 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21.41 on 28 degrees of freedom
## Multiple R-squared: 0.9349, Adjusted R-squared: 0.9326
## F-statistic: 402.3 on 1 and 28 DF, p-value: < 2.2e-16
The new on base, new slug and sum of on base and slug were all better predictors of runs that the old variables. They have an R\(^2\) value of .8491, .8969 and .9349, respectively. The best predictor of runs is new on base and slug together. 93.49% of the variability in runs is accounted for by being new on base and slugging.
Testing Linearity
plot(basslug$residuals ~ mlb11$new_obs)
abline(h = 0, lty = 3)
There is no pattern in the residuals. The relationship between being new on base and slugging and runs appears to be linear.
Testing for Nearly Normal Residuals
hist(basslug$residuals)
The residuals look nearly normal.
Testing Constant Variability
Looking at the graph of Runs vs. On Base and Slugging Percentage, one can see that the variability of the data is consistent.
All of the tests for using a linear regression model are satisfied.
This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel from a lab written by the faculty and TAs of UCLA Statistics.